Guidelines for code-golf on SO
The shortest code by character count to display a representation of a circle of radius R
using the *
character, followed by an approximation of π.
Input is a single number, R
.
Since most computers seem to have almost 2:1 ratio you should only output lines where y
is odd. This means that when R
is odd you should print R-1
lines. There is a new testcase for R=13
to clarify.
eg.
Input
5
Output Correct Incorrect
3 ******* 4 *******
1 ********* 2 *********
-1 ********* 0 ***********
-3 ******* -2 *********
2.56 -4 *******
3.44
Edit: Due to widespread confusion caused by odd values of R
, any solutions that pass the 4 test cases given below will be accepted
The approximation of π is given by dividing twice the number of *
characters by R²
.
The approximation should be correct to at least 6 significant digits.
Leading or trailing zeros are permitted, so for example any of 3
, 3.000000
, 003
is accepted for the inputs of 2
and 4
.
Code count includes input/output (i.e., full program).
Input
2
Output
***
***
3.0
Input
4
Output
*****
*******
*******
*****
3.0
Input
8
Output
*******
*************
***************
***************
***************
***************
*************
*******
3.125
Input
10
Output
*********
***************
*****************
*******************
*******************
*******************
*******************
*****************
***************
*********
3.16
Input
13
Output
*************
*******************
*********************
***********************
*************************
*************************
*************************
*************************
***********************
*********************
*******************
*************
2.98224852071
Just in case, I am using OpenBSD and some supposedly non-portable extensions at this point.
93 chars. This is based on same formula as FORTRAN solution (slightly different results than test cases). Calculates X^2=R^2-Y^2 for every Y
[rdPr1-d0<p]sp1?dsMdd*sRd2%--
[dd*lRr-vddlMr-32rlpxRR42r2*lpxRRAP4*2+lN+sN2+dlM>y]
dsyx5klNlR/p
88 chars. Iterative solution. Matches test cases. For every X and Y checks if X^2+Y^2<=R^2
1?dsMdd*sRd2%--sY[0lM-[dd*lYd*+lRr(2*d5*32+PlN+sN1+dlM!<x]dsxxAPlY2+dsYlM>y]
dsyx5klNlR/p
To run dc pi.dc
.
Here is an older annotated version:
# Routines to print '*' or ' '. If '*', increase the counter by 2
[lN2+sN42P]s1
[32P]s2
# do 1 row
# keeping I in the stack
[
# X in the stack
# Calculate X^2+Y^2 (leave a copy of X)
dd*lYd*+
#Calculate X^2+Y^2-R^2...
lR-d
# .. if <0, execute routine 1 (print '*')
0>1
# .. else execute routine 2 (print ' ')
0!>2
# increment X..
1+
# and check if done with line (if not done, recurse)
d lM!<x
]sx
# Routine to cycle for the columns
# Y is on the stack
[
# push -X
0lM-
# Do row
lxx
# Print EOL
10P
# Increment Y and save it, leaving 2 copies
lY 2+ dsY
# Check for stop condition
lM >y
]sy
# main loop
# Push Input value
[Input:]n?
# Initialize registers
# M=rows
d sM
# Y=1-(M-(M%2))
dd2%-1r-sY
# R=M^2
d*sR
# N=0
0sN
[Output:]p
# Main routine
lyx
# Print value of PI, N/R
5klNlR/p